Is the homomorphism $mathbbQGto prod M_chi_i(1)(mathbbQ)$ given by $x mapsto (rho_i(x))_i$ an isomorphism? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Class function as a characterSum of squares of dimensions of irreducible characters.faithful representation related to the centerRepresentation theory and character proof$chi(g)$ group character $Rightarrow$ $chi(g^m)$ group characterOccurrences of trivial representation is equal to dimension of $vin V:varphi(g)v=v$.Why do the characters of an abelian group form a group?characters and representations of extra-special $p$-groupsIf $T$ is an algebraic torus, is there a difference between $operatornameIrr(T)$ and $X(T)$?The ring $R (G)$ in Serre's Linear Representations of Finite Groups

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Is the homomorphism $mathbbQGto prod M_chi_i(1)(mathbbQ)$ given by $x mapsto (rho_i(x))_i$ an isomorphism?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Class function as a characterSum of squares of dimensions of irreducible characters.faithful representation related to the centerRepresentation theory and character proof$chi(g)$ group character $Rightarrow$ $chi(g^m)$ group characterOccurrences of trivial representation is equal to dimension of $vin V:varphi(g)v=v$.Why do the characters of an abelian group form a group?characters and representations of extra-special $p$-groupsIf $T$ is an algebraic torus, is there a difference between $operatornameIrr(T)$ and $X(T)$?The ring $R (G)$ in Serre's Linear Representations of Finite Groups










2












$begingroup$


If we have the group algebra $mathbbQG$ and $chi_1,...,chi_n$ the irreducible characters of $G$ afforded by the representation $rho_1,...,rho_n$, is it true that the map:



$mathbbQGto prod M_chi_i(1)(mathbbQ)$ sending $xin mathbbQG$ to $(rho_i(x))_i$



is an isomorphism?



We know that $mathbbQG$ can be decomposed into simple algebras, and every simple algebra produces an irreducible character. Can I mix these facts to give a positive answer to the question?










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    If we have the group algebra $mathbbQG$ and $chi_1,...,chi_n$ the irreducible characters of $G$ afforded by the representation $rho_1,...,rho_n$, is it true that the map:



    $mathbbQGto prod M_chi_i(1)(mathbbQ)$ sending $xin mathbbQG$ to $(rho_i(x))_i$



    is an isomorphism?



    We know that $mathbbQG$ can be decomposed into simple algebras, and every simple algebra produces an irreducible character. Can I mix these facts to give a positive answer to the question?










    share|cite|improve this question











    $endgroup$














      2












      2








      2


      1



      $begingroup$


      If we have the group algebra $mathbbQG$ and $chi_1,...,chi_n$ the irreducible characters of $G$ afforded by the representation $rho_1,...,rho_n$, is it true that the map:



      $mathbbQGto prod M_chi_i(1)(mathbbQ)$ sending $xin mathbbQG$ to $(rho_i(x))_i$



      is an isomorphism?



      We know that $mathbbQG$ can be decomposed into simple algebras, and every simple algebra produces an irreducible character. Can I mix these facts to give a positive answer to the question?










      share|cite|improve this question











      $endgroup$




      If we have the group algebra $mathbbQG$ and $chi_1,...,chi_n$ the irreducible characters of $G$ afforded by the representation $rho_1,...,rho_n$, is it true that the map:



      $mathbbQGto prod M_chi_i(1)(mathbbQ)$ sending $xin mathbbQG$ to $(rho_i(x))_i$



      is an isomorphism?



      We know that $mathbbQG$ can be decomposed into simple algebras, and every simple algebra produces an irreducible character. Can I mix these facts to give a positive answer to the question?







      abstract-algebra group-theory ring-theory representation-theory characters






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 24 at 16:29









      Brahadeesh

      6,59152465




      6,59152465










      asked Mar 24 at 14:26









      AlopisoAlopiso

      1379




      1379




















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          No. For a very simple example, let $G$ be cyclic of order 3. Then $G$ has only two irreducible representations over $mathbbQ$: the trivial representation $rho_1$, and the quotient $rho_2$ of the regular representation by the trivial representation (which is an irreducible representation of degree $2$). Your map $mathbbQGto M_1(mathbbQ)times M_2(mathbbQ)$ then cannot be surjective, since $mathbbQG$ is $3$-dimensional and $M_1(mathbbQ)times M_2(mathbbQ)$ is $5$-dimensional. What's going on here is that the subring generated by the image of $rho_2$ is not the full matrix ring $M_2(mathbbQ)$; instead it's a $2$-dimensional subring which is isomorphic to the field $mathbbQ(zeta)$ where $zeta$ is a primitive cube root of $1$.



          In general, the image of $rho_i$ is a matrix ring over the endomorphism ring $D_i$ of $rho_i$. This endomorphism ring $D_i$ is a division algebra since $rho_i$ is irreducible. If $D_i$ is just $mathbbQ$, then the image of $rho_i$ will be all of $M_chi_i(1)(mathbbQ)$, but if $D_i$ is larger than $mathbbQ$ then the image of $rho_i$ is a proper subring of $M_chi_i(1)(mathbbQ)$ (namely, the subring of elements that commute with every element of $D_i$).



          Your statement would be correct if you were working over $mathbbC$ instead of $mathbbQ$. Over $mathbbC$, there are no finite-dimensional division algebras besides $mathbbC$ itself, so the image of every irreducible representation is the full matrix ring $M_chi_i(1)(mathbbC)$. In general, the Artin-Wedderburn theorem says a semisimple ring is isomorphic to the product of the images of its irreducible representations, which are matrix rings over the endomorphism rings of those representations (and those endomorphism rings are division algebras).






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thanks for your time, you help me a lot :D
            $endgroup$
            – Alopiso
            Mar 24 at 16:08


















          2












          $begingroup$

          This is not true. The group algebra of the cyclic group of order three is commutative, and decomposes as $mathbbQC_3cong mathbbQoplusmathbbQ(omega)$, where $omega$ is a primitive third root of unity.



          If you replace $mathbbQ$ with $mathbbC$, then the statement is true (in the latter case, $mathbbCC_3cong mathbbCoplusmathbbCoplusmathbbC$).






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Why is true for $mathbbQS_3 cong mathbbQoplus mathbbQoplus M_2(mathbbQ)$?
            $endgroup$
            – Alopiso
            Mar 24 at 15:49










          • $begingroup$
            $mathbbQC_3cong mathbbQ[x]/(x^3-1)$
            $endgroup$
            – David Hill
            Mar 24 at 16:11











          Your Answer








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          2 Answers
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          2 Answers
          2






          active

          oldest

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          active

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          votes






          active

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          5












          $begingroup$

          No. For a very simple example, let $G$ be cyclic of order 3. Then $G$ has only two irreducible representations over $mathbbQ$: the trivial representation $rho_1$, and the quotient $rho_2$ of the regular representation by the trivial representation (which is an irreducible representation of degree $2$). Your map $mathbbQGto M_1(mathbbQ)times M_2(mathbbQ)$ then cannot be surjective, since $mathbbQG$ is $3$-dimensional and $M_1(mathbbQ)times M_2(mathbbQ)$ is $5$-dimensional. What's going on here is that the subring generated by the image of $rho_2$ is not the full matrix ring $M_2(mathbbQ)$; instead it's a $2$-dimensional subring which is isomorphic to the field $mathbbQ(zeta)$ where $zeta$ is a primitive cube root of $1$.



          In general, the image of $rho_i$ is a matrix ring over the endomorphism ring $D_i$ of $rho_i$. This endomorphism ring $D_i$ is a division algebra since $rho_i$ is irreducible. If $D_i$ is just $mathbbQ$, then the image of $rho_i$ will be all of $M_chi_i(1)(mathbbQ)$, but if $D_i$ is larger than $mathbbQ$ then the image of $rho_i$ is a proper subring of $M_chi_i(1)(mathbbQ)$ (namely, the subring of elements that commute with every element of $D_i$).



          Your statement would be correct if you were working over $mathbbC$ instead of $mathbbQ$. Over $mathbbC$, there are no finite-dimensional division algebras besides $mathbbC$ itself, so the image of every irreducible representation is the full matrix ring $M_chi_i(1)(mathbbC)$. In general, the Artin-Wedderburn theorem says a semisimple ring is isomorphic to the product of the images of its irreducible representations, which are matrix rings over the endomorphism rings of those representations (and those endomorphism rings are division algebras).






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thanks for your time, you help me a lot :D
            $endgroup$
            – Alopiso
            Mar 24 at 16:08















          5












          $begingroup$

          No. For a very simple example, let $G$ be cyclic of order 3. Then $G$ has only two irreducible representations over $mathbbQ$: the trivial representation $rho_1$, and the quotient $rho_2$ of the regular representation by the trivial representation (which is an irreducible representation of degree $2$). Your map $mathbbQGto M_1(mathbbQ)times M_2(mathbbQ)$ then cannot be surjective, since $mathbbQG$ is $3$-dimensional and $M_1(mathbbQ)times M_2(mathbbQ)$ is $5$-dimensional. What's going on here is that the subring generated by the image of $rho_2$ is not the full matrix ring $M_2(mathbbQ)$; instead it's a $2$-dimensional subring which is isomorphic to the field $mathbbQ(zeta)$ where $zeta$ is a primitive cube root of $1$.



          In general, the image of $rho_i$ is a matrix ring over the endomorphism ring $D_i$ of $rho_i$. This endomorphism ring $D_i$ is a division algebra since $rho_i$ is irreducible. If $D_i$ is just $mathbbQ$, then the image of $rho_i$ will be all of $M_chi_i(1)(mathbbQ)$, but if $D_i$ is larger than $mathbbQ$ then the image of $rho_i$ is a proper subring of $M_chi_i(1)(mathbbQ)$ (namely, the subring of elements that commute with every element of $D_i$).



          Your statement would be correct if you were working over $mathbbC$ instead of $mathbbQ$. Over $mathbbC$, there are no finite-dimensional division algebras besides $mathbbC$ itself, so the image of every irreducible representation is the full matrix ring $M_chi_i(1)(mathbbC)$. In general, the Artin-Wedderburn theorem says a semisimple ring is isomorphic to the product of the images of its irreducible representations, which are matrix rings over the endomorphism rings of those representations (and those endomorphism rings are division algebras).






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thanks for your time, you help me a lot :D
            $endgroup$
            – Alopiso
            Mar 24 at 16:08













          5












          5








          5





          $begingroup$

          No. For a very simple example, let $G$ be cyclic of order 3. Then $G$ has only two irreducible representations over $mathbbQ$: the trivial representation $rho_1$, and the quotient $rho_2$ of the regular representation by the trivial representation (which is an irreducible representation of degree $2$). Your map $mathbbQGto M_1(mathbbQ)times M_2(mathbbQ)$ then cannot be surjective, since $mathbbQG$ is $3$-dimensional and $M_1(mathbbQ)times M_2(mathbbQ)$ is $5$-dimensional. What's going on here is that the subring generated by the image of $rho_2$ is not the full matrix ring $M_2(mathbbQ)$; instead it's a $2$-dimensional subring which is isomorphic to the field $mathbbQ(zeta)$ where $zeta$ is a primitive cube root of $1$.



          In general, the image of $rho_i$ is a matrix ring over the endomorphism ring $D_i$ of $rho_i$. This endomorphism ring $D_i$ is a division algebra since $rho_i$ is irreducible. If $D_i$ is just $mathbbQ$, then the image of $rho_i$ will be all of $M_chi_i(1)(mathbbQ)$, but if $D_i$ is larger than $mathbbQ$ then the image of $rho_i$ is a proper subring of $M_chi_i(1)(mathbbQ)$ (namely, the subring of elements that commute with every element of $D_i$).



          Your statement would be correct if you were working over $mathbbC$ instead of $mathbbQ$. Over $mathbbC$, there are no finite-dimensional division algebras besides $mathbbC$ itself, so the image of every irreducible representation is the full matrix ring $M_chi_i(1)(mathbbC)$. In general, the Artin-Wedderburn theorem says a semisimple ring is isomorphic to the product of the images of its irreducible representations, which are matrix rings over the endomorphism rings of those representations (and those endomorphism rings are division algebras).






          share|cite|improve this answer











          $endgroup$



          No. For a very simple example, let $G$ be cyclic of order 3. Then $G$ has only two irreducible representations over $mathbbQ$: the trivial representation $rho_1$, and the quotient $rho_2$ of the regular representation by the trivial representation (which is an irreducible representation of degree $2$). Your map $mathbbQGto M_1(mathbbQ)times M_2(mathbbQ)$ then cannot be surjective, since $mathbbQG$ is $3$-dimensional and $M_1(mathbbQ)times M_2(mathbbQ)$ is $5$-dimensional. What's going on here is that the subring generated by the image of $rho_2$ is not the full matrix ring $M_2(mathbbQ)$; instead it's a $2$-dimensional subring which is isomorphic to the field $mathbbQ(zeta)$ where $zeta$ is a primitive cube root of $1$.



          In general, the image of $rho_i$ is a matrix ring over the endomorphism ring $D_i$ of $rho_i$. This endomorphism ring $D_i$ is a division algebra since $rho_i$ is irreducible. If $D_i$ is just $mathbbQ$, then the image of $rho_i$ will be all of $M_chi_i(1)(mathbbQ)$, but if $D_i$ is larger than $mathbbQ$ then the image of $rho_i$ is a proper subring of $M_chi_i(1)(mathbbQ)$ (namely, the subring of elements that commute with every element of $D_i$).



          Your statement would be correct if you were working over $mathbbC$ instead of $mathbbQ$. Over $mathbbC$, there are no finite-dimensional division algebras besides $mathbbC$ itself, so the image of every irreducible representation is the full matrix ring $M_chi_i(1)(mathbbC)$. In general, the Artin-Wedderburn theorem says a semisimple ring is isomorphic to the product of the images of its irreducible representations, which are matrix rings over the endomorphism rings of those representations (and those endomorphism rings are division algebras).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 24 at 16:06

























          answered Mar 24 at 15:52









          Eric WofseyEric Wofsey

          193k14222353




          193k14222353











          • $begingroup$
            Thanks for your time, you help me a lot :D
            $endgroup$
            – Alopiso
            Mar 24 at 16:08
















          • $begingroup$
            Thanks for your time, you help me a lot :D
            $endgroup$
            – Alopiso
            Mar 24 at 16:08















          $begingroup$
          Thanks for your time, you help me a lot :D
          $endgroup$
          – Alopiso
          Mar 24 at 16:08




          $begingroup$
          Thanks for your time, you help me a lot :D
          $endgroup$
          – Alopiso
          Mar 24 at 16:08











          2












          $begingroup$

          This is not true. The group algebra of the cyclic group of order three is commutative, and decomposes as $mathbbQC_3cong mathbbQoplusmathbbQ(omega)$, where $omega$ is a primitive third root of unity.



          If you replace $mathbbQ$ with $mathbbC$, then the statement is true (in the latter case, $mathbbCC_3cong mathbbCoplusmathbbCoplusmathbbC$).






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Why is true for $mathbbQS_3 cong mathbbQoplus mathbbQoplus M_2(mathbbQ)$?
            $endgroup$
            – Alopiso
            Mar 24 at 15:49










          • $begingroup$
            $mathbbQC_3cong mathbbQ[x]/(x^3-1)$
            $endgroup$
            – David Hill
            Mar 24 at 16:11















          2












          $begingroup$

          This is not true. The group algebra of the cyclic group of order three is commutative, and decomposes as $mathbbQC_3cong mathbbQoplusmathbbQ(omega)$, where $omega$ is a primitive third root of unity.



          If you replace $mathbbQ$ with $mathbbC$, then the statement is true (in the latter case, $mathbbCC_3cong mathbbCoplusmathbbCoplusmathbbC$).






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Why is true for $mathbbQS_3 cong mathbbQoplus mathbbQoplus M_2(mathbbQ)$?
            $endgroup$
            – Alopiso
            Mar 24 at 15:49










          • $begingroup$
            $mathbbQC_3cong mathbbQ[x]/(x^3-1)$
            $endgroup$
            – David Hill
            Mar 24 at 16:11













          2












          2








          2





          $begingroup$

          This is not true. The group algebra of the cyclic group of order three is commutative, and decomposes as $mathbbQC_3cong mathbbQoplusmathbbQ(omega)$, where $omega$ is a primitive third root of unity.



          If you replace $mathbbQ$ with $mathbbC$, then the statement is true (in the latter case, $mathbbCC_3cong mathbbCoplusmathbbCoplusmathbbC$).






          share|cite|improve this answer









          $endgroup$



          This is not true. The group algebra of the cyclic group of order three is commutative, and decomposes as $mathbbQC_3cong mathbbQoplusmathbbQ(omega)$, where $omega$ is a primitive third root of unity.



          If you replace $mathbbQ$ with $mathbbC$, then the statement is true (in the latter case, $mathbbCC_3cong mathbbCoplusmathbbCoplusmathbbC$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 24 at 15:39









          David HillDavid Hill

          9,6561619




          9,6561619











          • $begingroup$
            Why is true for $mathbbQS_3 cong mathbbQoplus mathbbQoplus M_2(mathbbQ)$?
            $endgroup$
            – Alopiso
            Mar 24 at 15:49










          • $begingroup$
            $mathbbQC_3cong mathbbQ[x]/(x^3-1)$
            $endgroup$
            – David Hill
            Mar 24 at 16:11
















          • $begingroup$
            Why is true for $mathbbQS_3 cong mathbbQoplus mathbbQoplus M_2(mathbbQ)$?
            $endgroup$
            – Alopiso
            Mar 24 at 15:49










          • $begingroup$
            $mathbbQC_3cong mathbbQ[x]/(x^3-1)$
            $endgroup$
            – David Hill
            Mar 24 at 16:11















          $begingroup$
          Why is true for $mathbbQS_3 cong mathbbQoplus mathbbQoplus M_2(mathbbQ)$?
          $endgroup$
          – Alopiso
          Mar 24 at 15:49




          $begingroup$
          Why is true for $mathbbQS_3 cong mathbbQoplus mathbbQoplus M_2(mathbbQ)$?
          $endgroup$
          – Alopiso
          Mar 24 at 15:49












          $begingroup$
          $mathbbQC_3cong mathbbQ[x]/(x^3-1)$
          $endgroup$
          – David Hill
          Mar 24 at 16:11




          $begingroup$
          $mathbbQC_3cong mathbbQ[x]/(x^3-1)$
          $endgroup$
          – David Hill
          Mar 24 at 16:11

















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Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029